Integrand size = 18, antiderivative size = 60 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x} \, dx=a c d x+b c d x \text {arctanh}(c x)+a d \log (x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )-\frac {1}{2} b d \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b d \operatorname {PolyLog}(2,c x) \]
a*c*d*x+b*c*d*x*arctanh(c*x)+a*d*ln(x)+1/2*b*d*ln(-c^2*x^2+1)-1/2*b*d*poly log(2,-c*x)+1/2*b*d*polylog(2,c*x)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x} \, dx=a c d x+b c d x \text {arctanh}(c x)+a d \log (x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )+\frac {1}{2} b d (-\operatorname {PolyLog}(2,-c x)+\operatorname {PolyLog}(2,c x)) \]
a*c*d*x + b*c*d*x*ArcTanh[c*x] + a*d*Log[x] + (b*d*Log[1 - c^2*x^2])/2 + ( b*d*(-PolyLog[2, -(c*x)] + PolyLog[2, c*x]))/2
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6502, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(c d x+d) (a+b \text {arctanh}(c x))}{x} \, dx\) |
| \(\Big \downarrow \) 6502 |
| \(\displaystyle \int \left (c d (a+b \text {arctanh}(c x))+\frac {d (a+b \text {arctanh}(c x))}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a c d x+a d \log (x)+b c d x \text {arctanh}(c x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )-\frac {1}{2} b d \operatorname {PolyLog}(2,-c x)+\frac {1}{2} b d \operatorname {PolyLog}(2,c x)\) |
a*c*d*x + b*c*d*x*ArcTanh[c*x] + a*d*Log[x] + (b*d*Log[1 - c^2*x^2])/2 - ( b*d*PolyLog[2, -(c*x)])/2 + (b*d*PolyLog[2, c*x])/2
3.1.5.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e _.)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p, ( f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Time = 0.43 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22
| method | result | size |
| parts | \(a d \left (c x +\ln \left (x \right )\right )+b d \left (\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}\right )\) | \(73\) |
| derivativedivides | \(a d \left (c x +\ln \left (c x \right )\right )+b d \left (\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}\right )\) | \(75\) |
| default | \(a d \left (c x +\ln \left (c x \right )\right )+b d \left (\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x +1\right )}{2}-\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {\operatorname {dilog}\left (c x \right )}{2}\right )\) | \(75\) |
| risch | \(-\frac {\ln \left (-c x +1\right ) b c d x}{2}+a c d x +\frac {\ln \left (-c x +1\right ) b d}{2}+\ln \left (-c x \right ) a d +\frac {\operatorname {dilog}\left (-c x +1\right ) b d}{2}-a d -b d +\frac {\ln \left (c x +1\right ) b c d x}{2}+\frac {\ln \left (c x +1\right ) b d}{2}-\frac {\operatorname {dilog}\left (c x +1\right ) b d}{2}\) | \(90\) |
a*d*(c*x+ln(x))+b*d*(ln(c*x)*arctanh(c*x)+c*x*arctanh(c*x)+1/2*ln(c*x-1)+1 /2*ln(c*x+1)-1/2*dilog(c*x+1)-1/2*ln(c*x)*ln(c*x+1)-1/2*dilog(c*x))
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x} \, dx=\int { \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x} \,d x } \]
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x} \, dx=d \left (\int a c\, dx + \int \frac {a}{x}\, dx + \int b c \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \]
d*(Integral(a*c, x) + Integral(a/x, x) + Integral(b*c*atanh(c*x), x) + Int egral(b*atanh(c*x)/x, x))
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x} \, dx=\int { \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x} \,d x } \]
a*c*d*x + 1/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*d + 1/2*b*d*integ rate((log(c*x + 1) - log(-c*x + 1))/x, x) + a*d*log(x)
\[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x} \, dx=\int { \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x} \,d x } \]
Timed out. \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x} \, dx=\int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+c\,d\,x\right )}{x} \,d x \]